Datasheet

Data Sheet ADP1850
Rev. A | Page 21 of 32
LOOP COMPENSATION (SINGLE PHASE
OPERATION)
As with most current mode step-down controller, a transcon-
ductance error amplifier is used to stabilize the external voltage
loop. Compensating the ADP1850 is fairly easy; an RC compen-
sator is needed between COMPx and AGND. Figure 33 shows
the configuration of the compensation components: R
COMP
,
C
COMP
, and C
C2
. Because C
C2
is very small compared to C
COMP
,
to simplify calculation, C
C2
is ignored for the stability
compensation analysis.
ADP1850
FBx
C
COMP
G
m
0.6V
COMPx
AGND
R
COMP
C
C2
09440-034
Figure 33. Compensation Components
The open loop gain transfer function at angular frequency, s, is
given by
)()()( sZsZ
V
V
GGsH
FILTER
COMP
OUT
REF
CS
m
××××=
(1)
where:
G
m
is the transconductance of the error amplifier, 500 µS.
G
CS
is the tranconductance of the power stage.
Z
COMP
is the impedance of the compensation network.
Z
FILTER
is the impedance of the output filter.
V
REF
= 0.6 V.
G
CS
with units of A/V is given by
MINDSONCS
CS
RA
G
_
1
×
=
(2)
where:
A
CS
is the current sense gain of either 3 V/V, 6 V/V, 12 V/V, or
24 V/V set by the gain resistor between DLx and PGNDx.
R
DSON_MIN
is the low-side MOSFET minimum on resistance.
If a sense resistor, R
S
, is added in series with the low-side FET,
then G
CS
becomes
)(
1
_ SMINDSONCS
CS
RRA
G
+×
=
Because the zero produced by the ESR of the output capacitor is
not needed to stabilize the control loop, assuming ESR is small
the ESR is ignored for analysis. Then Z
FILTER
is given by
OUT
FILTER
sC
Z
1
=
(3)
Because C
C2
is small relative to C
COMP
, Z
COMP
can be simplified to
COMP
COMPCOMP
COMP
COMPCOMP
sC
CsR
sC
RZ
×+
=+=
1
1
(4)
At the crossover frequency, the open-loop transfer function is
unity or 0 dB, H (f
CROSS
) = 1. Combining Equation 1 and
Equation 3, Z
COMP
at the crossover frequency can be written as
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×π
=
REF
OUTOUT
CS
m
CROSS
CROSSCOMP
V
VC
GG
f
fZ
2
)(
(5)
The zero produced by R
COMP
and C
COMP
is
COMPCOMP
ZERO
CR
f
×π
=
2
1
(6)
At the crossover frequency, Equation 4 can be shown as
CROSS
ZERO
CROSS
COMPCROSSCOMP
f
ff
RfZ
2
)(
2
+
×=
(7)
Combining Equation 5 and Equation 7 and solving for
R
COMP
gives
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×π
×
+
=
REF
OUTOUT
CS
m
CROSS
ZERO
CROSS
CROSS
COMP
V
VC
GG
f
ff
f
R
2
22
(8)
Choose the crossover and zero frequencies as follows:
12
SW
CROSS
f
f
=
(9)
484
SWCROSS
ZERO
ff
f
==
(10)
Substituting Equation 2, Equation 9, and Equation 10 into
Equation 8 yields
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×π
××=
REF
OUTOUT
m
CROSS
DSONCSCOMP
V
VC
G
f
RAR
2
97.0
(11)
where:
G
m
is the transconductance of the error amplifier, 500 µS.
A
CS
is the current sense gain of 3 V/V, 6 V/V, 12 V/V, or 24 V/V.
R
DSON
is on resistance of the low-side MOSFET.
V
REF
= 0.6 V.
And combining Equation 6 and Equation 10 yields
CROSSCOMP
COMP
fR
C
×π
=
2
(12)
Note that the previous simplified compensation equations for
R
COMP
and C
COMP
yield reasonable results in f
CROSS
and phase
margin assuming that the compensation ramp current is ideal.
Varying the ramp current or deviating the ramp current from
ideal can affect f
CROSS
and phase margin.
And lastly, set C
C2
to
COMPCCOMP
CCC ×≤≤×
10
1
20
1
2
(13)